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PS6_Solutions

# PS6_Solutions - AEM 4150 PRICE ANALYSIS Fall 2008 Problem...

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AEM 4150 PRICE ANALYSIS Fall 2008 Problem Set #6 Suggested Solutions 1. (a) Farm-processor marketing margin = processor price – farm price = \$1.45 – 1 = \$.45. Processor-retail marketing margin = retail price – processor price = \$3.20 – 1.45 = 1.75 Farm-retail marketing margin = retail price – farm price = \$3.20 – 1 = \$2.20 (b) Wegmans’ profit = processor-retail marketing margin – cost = \$1.75 – .6 = \$1.15. Yes. Marketing margin is the price of marketing services. Profit is the return to entrepreneurship and therefore is a component of marketing margin. (c) k = 0.8. MM = Pf – (1/k)*pf = \$3.20 – (1/0.8)*1 = \$1.95 2. (a) Graph Dr and Sf in the following diagram. P f *=23.3 D f : Pf=60-Q Q*=36.7 P r * =63.3 S f : Pf = -50+2Q S r : Pr=-10+2Q D r : Pr=100-Q Q P (b) Derived Demand facing growers (Df )= primary demand – MM = 100 – Q – 40 = 60 – Q P f = 60 – Q 1

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Derived Retail Supply Function (S r ) = primary supply + MM = –50 + 2Q + 40 = –10 + 2Q P r = -10 + 2Q Plot Df and Sr in the above diagram. (c) At the retail level, set S r = D r and solve for Q* –10+2Q =100-Q 3Q=110 Q* = 36.7 Check: Set S f = D f -50+2Q = 60-Q 3Q = 110 Q* = 36.7 P r * = 100 – 36.7 P r * = 63.3 P f * = 60 – 36.7
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